A student has 25 mL of 6M HCl. What is the minimum volume of 8% NaOH (w/v) required to neutralize the acid?

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#1

A student has 25 mL of 6M HCl. What is the minimum volume of 8% NaOH (w/v) required to neutralize the acid?

Answer:

6(M/L)*0.025L=. 15 moles of HCL
1 L ∼ 1000g → 8% NaOH= 80g/1000g = 80g/L
→NaOH molecular weight = 40 g/mole
80(g/L) * 1/40 (mol/g) = 2 (mol/L)
(6)(0.025)= (2)(x)
0.15=(2)(x)
x= 0.075L = 75ml